Solution: Given: First Plane: $2 x-y+4 z=5$ [On multiply both the sides by $2.5]$ We obtain, $5 x-2.5 y+10 z=12.5 \ldots$ Second Plane: $5 x-2.5 y+10 z=6 \ldots$ Therefore, $\begin{array}{l}...
Solution: It is known to us that the distance between two parallel planes $A x+B y+C z=d_{1}$ and $A x+B y+C z=d_{2}$ is given as...
Solution: It is known to us that the distance of the point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ from the plane $\mathrm{Ax}+\mathrm{By}+\mathrm{Cz}$ $=\mathrm{D}$ is given...
and 
Solution: The vector eq. of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\overrightarrow{\mathrm{b}}$ is...
and parallel to the planes 
and 
Solution: The vector eq. of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is $\vec{r}=\vec{a}+\lambda \vec{b}$ It is given that the line passes...
from the point of intersection of the line 
and the plane 
Solution: It is given that, The eq. of line is $\overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})...
and 
And which is perpendicular to the plane 
Solution: It is known that, The eq. of any plane through the line of intersection of the planes $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}_{1}}=\mathrm{d}_{1}$ and...
Solution: It is known to us that the eq. of a plane passing through $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and perpendicular to a line with direction ratios $A, B, C$ is given...
and 
and parallel to 
-axis.Solution: It is known to us that, The eq. of any plane through the line of intersection of the planes $\vec{r} \cdot \overrightarrow{n_{1}}=d_{1}$ and $\vec{r} \cdot \overrightarrow{n_{2}}=d_{2}$ is...
, then find the value of 
.Solution: It is known to us that the distance of a point with position vector $\vec{a}$ from the plane $\vec{r} \cdot \vec{n}=d$ is given as $\left|\frac{\vec{a} \cdot \vec{n}-d}{|\vec{n}|}\right|$...
Solution: It is known to us that the eq. of a plane passing through $\left(x_{1}, y_{1}, z_{1}\right)$ is given by $A\left(x-x_{1}\right)+B\left(y-y_{1}\right)+C\left(z-z_{1}\right)=0$ Where, A, B,...
Solution: It is known to us that the eq. of a line passing through two points $A\left(x_{1}, y_{1}, z_{1}\right)$ and $B\left(x_{2}, y_{2}, z_{2}\right)$ is given as...
Solution: It is known to us that the vector eq. of a line passing through two points with position vectors $\vec{a}$ and $b$ is given as...
Solution: It is known to us that the vector eq. of a line passing through two points with position vectors $\vec{a}$ and $\vec{b}$ is given as...
Solution: It is known to us that the shortest distance between lines with vector equations $\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$ and $\vec{r}=\overrightarrow{a_{2}}+\lambda...
Solution: The eq. of a plane passing through $\left(x_{1}, y_{1}, z_{1}\right)$ and perpendicular to a line with direction ratios $A, B, C$ is given as...
Solution: The vector eq. of a line passing through a point with position vector $\vec{a}$ and parallel to vector $\vec{b}$ is given as...
are perpendicular, find the value of 
.Solution: It is known to us that the two lines $\frac{x-1}{3 k}=\frac{y-2}{1}=\frac{z-3}{-5} \text { and }$ $\frac{\mathrm{x}-1}{3 \mathrm{k}}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-3}{-5}$ are...
Solution: It is known to us that the angle between the lines with direction ratios $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ is given by $\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1}...
Solution: It is known to us that, eq. of a line passing through $\left(x_{1}, y_{1}, z_{1}\right)$ and parallel to a line with direction ratios $a, b, c$ is...
Solution: The angle between the lines with direction ratios $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ is given by $\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1}...
and 
are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are 
Solution: Let's consider $l, m, n$ be the direction cosines of the line perpendicular to each of the given lines. Therefore, $ll_{1}+m m_{1}+n n_{1}=0 \ldots(1)$ And...
Solution: Let's consider $O A$ be the line joining the origin $(0,0,0)$ and the point $A(2,1,1)$. And let $B C$ be the line joining the points $B(3,5,-1)$ and $C(4,3,-1)$ Therefore the direction...
Solution: (a) The length of perpendicular from the point $(2,3,-5)$ on the plane $x+2 y-2 z=9 \Rightarrow x+2 y-2 z-9=0$ is $\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$...
Solution: (a) The distance of the point $(0,0,0)$ from the plane $3 x-4 y+12=3 \Rightarrow$ $3 x-4 y+12 z-3=0$ is $\begin{array}{l} \frac{\left|a x_{1}+b y_{1}+c...
Solution: (a) $4 x+8 y+z-8=0$ and $y+z-4=0$ It is given that The eq. of the given planes are $4 x+8 y+z-8=0 \text { and } y+z-4=0$ It is known to us that, two planes are $\perp$ if the direction...
Solution: (a) $2 x-2 y+4 z+5=0$ and $3 x-3 y+6 z-1=0$ It is given that The eq. of the given planes are $2 x-2 y+4 z+5=0$ and $x-2 y+5=0$ It is known to us that, two planes are $\perp$ if the...
Solution: (a) $7 x+5 y+6 z+30=0$ and $3 x-y-10 z+4=0$ It is given that The eq. of the given planes are $7 x+5 y+6 z+30=0$ and $3 x-y-10 z+4=0$ Two planes are $\perp$ if the direction ratio of the...
Solution: It is given that The eq. of the given planes are $\vec{r}(2 \hat{i}+2 \hat{j}-3 \hat{k})=5 \text { and } \vec{r}(3 \hat{i}-3 \hat{j}+5 \hat{k})=5$ If $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$...
Solution: Let's say that the eq. of the plane that passes through the two-given planes $x+y+z=1$ and $2 x+3 y+4 z=5$ is $\begin{array}{l} (x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0 \\ (2 \lambda+1) x+(3...
and through the point 
Solution: Let's consider the vector eq. of the plane passing through the intersection of the planes are $\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})=7...
Solution: It is given that Eq. of the plane passes through the intersection of the plane is given by $(3 x-y+2 z-4)+\lambda(x+y+z-2)=0$ and the plane passes through the points $(2,2,1)$ Therefore,...
-axis and parallel to ZOX plane.Solution: It is known to us that the equation of the plane $\mathrm{ZOX}$ is $\mathrm{y}=0$ Therefore, the equation of plane parallel to $\mathrm{ZOX}$ is of the form, $\mathrm{y}=\mathrm{a}$ As the...
Solution: It is given that The plane $2 x+y-z=5$ Let us express the equation of the plane in intercept form $x / a+y / b+z / c=1$ Where $a, b, c$ are the intercepts cut-off by the plane at $x, y$...
Solution: (a) It is given that, The points are $(1,1,-1),(6,4,-5),(-4,-2,3)$. Let, $\begin{array}{l} =\left|\begin{array}{ccc} 1 & 1 & -1 \\ 6 & 4 & -5 \\ -4 & -2 & 3...
and the normal to the plane is 
and the normal vector to the plane is 
Solution: (a) That passes through the point $(1,0,-2)$ and the normal to the plane is $\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ Let's say that the position vector of the point $(1,0,-2)$...
Solution: (a) $x+y+z=1$ Let the coordinate of the foot of $\perp \mathrm{P}$ from the origin to the given plane be $P(x, y, z)$ $x+y+z=1$ The direction ratio are $(1,1,1)$ $\begin{array}{l}...
![\overrightarrow{\mathrm{r}} \cdot[(\mathrm{s}-2 \mathrm{t}) \hat{\mathrm{i}}+(3-\mathrm{t}) \hat{\mathrm{j}}+(2 \mathrm{~s}+\mathrm{t}) \hat{\mathrm{k}}]=15](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0789422b9ad2fd168094a22f1dd6043a_l3.png "Rendered by QuickLaTeX.com")
Solution: Let $\overrightarrow{\mathrm{r}}$ be the position vector of $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is given by $\overrightarrow{\mathrm{r}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y}...
Solution: (a) It is given that, The equation of the plane. Let $\overrightarrow{\mathrm{r}}$ be the position vector of $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is given by $\vec{r}=x...
Solution: It is given that, The vector $3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}$ Vector equation of the plane with position vector $\overrightarrow{\mathrm{r}}$ is $\vec{r} \cdot...
Solution: (a) $2 x+3 y-z=5$ It is given that The eq. of the plane, $2 x+3 y-z=5 \ldots$. (1) The direction ratio of the normal $(2,3,-1)$ Using the formula,...
Solution: (a) $z=2$ It is given that The eq. of the plane, $z=2$ or $0 x+0 y+z=2 \ldots (1) .$ The direction ratio of the normal $(0,0,1)$ Using the formula, $\begin{array}{l}...
Solution: Consider the given equations $\begin{array}{l} \Rightarrow \vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k} \\ \vec{r}=\hat{i}-t \hat{i}+t \hat{j}-2 \hat{j}+3 \hat{k}-2 t \hat{k} \\...
and 
Solution: It is known to us that shortest distance between two lines $\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$ and $\vec{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$...
and 
Solution: It is known to us that the shortest distance between two lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$ is given as:...
Solution: It is known to us that the shortest distance between two lines $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}_{1}}+\lambda \overrightarrow{\mathrm{b}_{1}}$ and...
and 
are perpendicular to each other.Solution: The equations of the given lines are $\frac{\mathrm{x}-5}{7}=\frac{\mathrm{y}+2}{-5}=\frac{\mathrm{z}}{1}$ and $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{3}$ Two lines...
and 
are at right angles.Solution: The standard form of a pair of Cartesian lines is:...
and 
and 
Solution: Let's consider $\theta$ be the angle between the given lines. If $\theta$ is the acute angle between $\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$ and...
Solution: It is given that Let's calculate the vector form: The vector eq. of as line which passes through two points whose position vectors are $\vec{a}$ and $\vec{b}$ is...
Solution: Given: The origin $(0,0,0)$ and the point $(5,-2,3)$ It is known to us that The vector eq. of as line which passes through two points whose position vectors are $\vec{a}$ and $\vec{b}$ is...
Write its vector form.Solution: It is given that The Cartesian equation is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} \ldots \text { (1) }$ It is known to us that The Cartesian eq. of a line passing through a point...
and parallel to the line given by 
Solution: It is given that The points $(-2,4,-5)$ It is known that Now, the Cartesian equation of a line through a point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and having...
and 
. is in the directionSolution: Given: Vector equation of a line that passes through a given point whose position vector is $\vec{a}$ and parallel to a given vector $\vec{b}$ is $\vec{r}=\vec{a}+\lambda \vec{b}$ Let,...
Solution: Given that, Line passes through the point $(1,2,3)$ and is parallel to the vector. It is known to us that Vector eq. of a line that passes through a given point whose position vector is...
Solution: The points $(4,7,8),(2,3,4)$ and $(-1,-2,1),(1,2,5)$. Consider $A B$ be the line joining the points, $(4,7,8),(2,3,4)$ and $C D$ be the line through the points $(-1,-2$, 1), $(1,2,5)$. So...
Solution: Given that The points $(1,-1,2),(3,4,-2)$ and $(0,3,2),(3,5,6)$. Let's consider $A B$ be the line joining the points, $(1,-1,2)$ and $(3,4,-2)$, and $C D$ be the line through the points...
Are mutually perpendicular.Solution: Consider the direction cosines of $L_{1}, L_{2}$ and $L_{3}$ be $l_{1}, m_{1}, n_{1} ; l_{2}, m_{2}, n_{2}$ and $l_{3}, m_{3}, n_{3}$. It is known that If $\mathrm{f}_{1}, \mathrm{~m}_{1},...
Solution: Given that, The vertices are $(3,5,-4),(-1,1,2)$ and $(-5,-5,-2)$. Firstly find the direction ratios of $\mathrm{AB}$ Where, $A=(3,5,-4)$ and $B=(-1,1,2)$ Ratio of $A...
Solution: If the direction ratios of two lines segments are proportional, then the lines are collinear. It is given that $\mathrm{A}(2,3,4), \mathrm{B}(-1,-2,1), \mathrm{C}(5,8,7)$ The direction...
Solution: Given that, The direction ratios are $-18,12,-4$ Where, $a=-18, b=12, c=-4$ Consider the direction ratios of the line as $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ Direction cosines are...
Solution: Given that, Angles are equal. Let the angles be $\alpha, \beta, \mathrm{Y}$ The direction cosines of the line be I, $\mathrm{m}$ and $\mathrm{n}$ $I=\cos \alpha, m=\cos \beta \text { and }...
Solution: Let's consider the direction cosines of the line be $I, m$ and $n$. Let $\alpha=90^{\circ}, \beta=135^{\circ}$ and $\mathrm{y}=45^{\circ}$ Therefore, $I=\cos \alpha, m=\cos \beta \text {...
![\[\left( \mathbf{3},\text{ }\mathbf{4},\text{ }\mathbf{5} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-caaf0c359ce8ed37268a1c6c3ead100a_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{1},\text{ }\mathbf{3},\text{ }\mathbf{7} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-500c94a3413872cbb0bc8e13c1d088d0_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{P}{{\mathbf{A}}^{\mathbf{2}}}~+\text{ }\mathbf{P}{{\mathbf{B}}^{\mathbf{2}}}~=\text{ }{{\mathbf{k}}^{\mathbf{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-48cce60e989713a6abb9ef8de8a49e52_l3.png "Rendered by QuickLaTeX.com")
Given: The points A \[\left( \mathbf{3},\text{ }\mathbf{4},\text{ }\mathbf{5} \right)\]and B \[\left( \mathbf{1},\text{ }\mathbf{3},\text{ }\mathbf{7} \right)\]...
![\[4\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3c0fa656ae249b0b1e03c8cd2e7be583_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{4} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-199d64e6806536c1b3cb857efe67f5ed_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{8},\text{ }\mathbf{0},\text{ }\mathbf{10} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-af725ac8ef019d8a7309e2821100c31a_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{k}:\text{ }\mathbf{1}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-518bc4432af641d05fe2a1affe87989c_l3.png "Rendered by QuickLaTeX.com")
Solution: Given: The coordinates of the points P \[\left( \mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{4} \right)\] and Q \[\left( \mathbf{8},\text{ }\mathbf{0},\text{ }\mathbf{10} \right)\]. Let...
![\[5\sqrt{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dce6450fc38db05cf3ea597cfff6b259_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{3},-\text{ }\mathbf{2},\text{ }\mathbf{5} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-817f357616717af1a3f7ba47dc39e134_l3.png "Rendered by QuickLaTeX.com")
Let the point on y-axis be A \[\left( 0,\text{ }y,\text{ }0 \right)\]. Then, it is given that the distance between the points A \[\left( 0,\text{ }y,\text{ }0 \right)\]and P \[\left( 3,-\text{...
![\[(\mathbf{2a},\text{ }\mathbf{2},\text{ }\mathbf{6})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-21601bb2fcbfbfc6a14be26b9355cbbf_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \text{ }\mathbf{4},\text{ }\mathbf{3b},\text{ }\mathbf{10} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-885e8872810191a196a77e9c3b732fa1_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{8},\text{ }\mathbf{14},\text{ }\mathbf{2c} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-94c3d49d7048725d17582e3162b1f873_l3.png "Rendered by QuickLaTeX.com")
Given: The vertices of the triangle are P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c). We know that the coordinates of the centroid of the triangle, whose vertices are \[({{x}_{1}},\text{...
![\[\left( \mathbf{0},\text{ }\mathbf{0},\text{ }\mathbf{6} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b4507c24b9833e46e57563cd7f680a4f_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{0},\text{ }\mathbf{4},\text{ }\mathbf{0} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cad57c36fd56e5ce3e723499ee15f4bc_l3.png "Rendered by QuickLaTeX.com")
![\[\left( 6,\text{ }0,\text{ }0 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0e0e30c6a01c24facaba8eab223cb371_l3.png "Rendered by QuickLaTeX.com")
According to the question: The vertices of the triangle are A \[\left( \mathbf{0},\text{ }\mathbf{0},\text{ }\mathbf{6} \right)\],B \[\left( \mathbf{0},\text{ }\mathbf{4},\text{ }\mathbf{0}...
![\[\left( \mathbf{3},\text{ }\text{ }\mathbf{1},\text{ }\mathbf{2} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aee8367f95fe0a1003f8bb7c21eaaa79_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{1},\text{ }\mathbf{2},\text{ }\text{ }\mathbf{4} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1ccf9b6a0ac05765abdbd55e4a3cec65_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \text{ }\mathbf{1},\text{ }\mathbf{1},\text{ }\mathbf{2} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cc0f840718d8caac43b743f7466304e4_l3.png "Rendered by QuickLaTeX.com")
According to the question: ABCD is a parallelogram, with vertices A\[\left( \mathbf{3},\text{ }\text{ }\mathbf{1},\text{ }\mathbf{2} \right)\], B \[\left( \mathbf{1},\text{ }\mathbf{2},\text{...
![\[\left( \mathbf{4},\text{ }\mathbf{2},\text{ }\text{ }\mathbf{6} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7b2b319c83af0f4415dd9c78671ae2e7_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{10},\text{ }\mathbf{16},\text{ }\mathbf{6} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b9fc798b7f04b8225ee220ac04bcf0cd_l3.png "Rendered by QuickLaTeX.com")
Consider A \[({{x}_{1}},\text{ }{{y}_{1}},\text{ }{{z}_{1}})\]and B \[({{x}_{2}},\text{ }{{y}_{2}},\text{ }{{z}_{2}})\]trisect the line segment joining the points P \[\left( \mathbf{4},\text{...
![\[\left( \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{1} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dfc9b15e84c73dbb78d47bc9d461ffe2_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{0},\text{ }\mathbf{1}/\mathbf{3},\text{ }\mathbf{2} \right)~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0ebe890b071a03a5f762eef8bc17b94e_l3.png "Rendered by QuickLaTeX.com")
Consider the point P divides AB in the ratio \[k:\text{ }1\]. By using section formula, So we have, Now, we check if for some value of k, the point coincides with the point C. Put \[\left( -k+2...
![\[\left( -\mathbf{2},\text{ }\mathbf{4},\text{ }\mathbf{7} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8b71485cf20e82a2a5a637c3422e2109_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{3},-\text{ }\mathbf{5},\text{ }\mathbf{8} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-05e14e4f32602279f3ac57eb034e0e97_l3.png "Rendered by QuickLaTeX.com")
Solution: Let the line segment formed by joining the points A \[\left(-\mathbf{2},\text{ }\mathbf{4},\text{ }\mathbf{7} \right)\]and B \[\left( \mathbf{3},-\text{ }\mathbf{5},\text{ }\mathbf{8}...
![\[\left( \mathbf{3},\text{ }\mathbf{2},\text{ }\text{ }\mathbf{4} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0b789d716677da76a2f2575aa3db7ee5_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{5},\text{ }\mathbf{4},\text{ }\text{ }\mathbf{6} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bc05994728fe8c2319e9d7d4c55a3235_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{9},\text{ }\mathbf{8},\text{ }\mathbf{10} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1f5aaf3ddebde616540844806ed2797f_l3.png "Rendered by QuickLaTeX.com")
Solution: Let us consider B divides AC in the ratio \[k:\text{ }1\]. By using section formula, So, we have \[9k\text{ }+\text{ }3\text{ }=\text{ }5\text{ }\left( k+1 \right)\] \[9k\text{ }+\text{...
![\[\left( \text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{5} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0adda3fe0ac17f8aa03c8923cc60645f_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{1},\text{ }\text{ }\mathbf{4},\text{ }\mathbf{6} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-75d0720435105f1c6905635a5d83b506_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{2}:\text{ }\mathbf{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a499d63c12c0dcb11e01e065feb57607_l3.png "Rendered by QuickLaTeX.com")
Solution: Let the line segment joining the points A \[\left( \text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{5} \right)\]and B \[\left( \mathbf{1},\text{ }\text{ }\mathbf{4},\text{ }\mathbf{6}...
![\[\left( \mathbf{4},\text{ }\mathbf{0},\text{ }\mathbf{0} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-32f234dc0219023b5a8ec08e2f8ebe98_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \text{ }\mathbf{4},\text{ }\mathbf{0},\text{ }\mathbf{0} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f71743c7bd55639820fe3b8928c19eae_l3.png "Rendered by QuickLaTeX.com")
Let p \[\left( \mathbf{4},\text{ }\mathbf{0},\text{ }\mathbf{0} \right)\]& q \[\left( \text{ }\mathbf{4},\text{ }\mathbf{0},\text{ }\mathbf{0} \right)\] Let the coordinates of point A be (x, y,...
![\[\left( \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ac64e1aeb36ee1ed893917d3ce1279d7_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{3},\text{ }\mathbf{2},\text{ -}\mathbf{1} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0bce7b0f8491ecbff79a9dd73cee51d2_l3.png "Rendered by QuickLaTeX.com")
Solution: Let $A(1,2,3) \& B(3,2,-1)$ Let point $\mathrm{P}$ be $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ Since it is given that po i.e. $P A=P B$ Firstly let us calculate Calculating PA $P...
![\[\left( 1,\text{ }2,\text{ }1 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c8e408e3cdf2569a678936b313cac0e1_l3.png "Rendered by QuickLaTeX.com")
![\[\left( 1,\text{ }2,\text{ }5 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d5b56c33ecc790bf4c34e2d57d3f8cdb_l3.png "Rendered by QuickLaTeX.com")
![\[\left( 4,\text{ }7,\text{ }8 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0f3656ef03bff2d276c934983c5bb4c7_l3.png "Rendered by QuickLaTeX.com")
![\[\left( 2,\text{ }3,\text{ }4 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-01cb19d575963567635f5f0b0cb97398_l3.png "Rendered by QuickLaTeX.com")
Let the points be: p\[\left( 1,\text{ }2,\text{ }1 \right)\], q\[\left( 1,\text{ }2,\text{ }5 \right)\],r\[\left( 4,\text{ }7,\text{ }8 \right)\] & s\[\left( 2,\text{ }3,\text{ }4 \right)\] pqrs...
![\[\left( 0,\text{ }7,\text{ }10 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-80f5b6e706c16a87a4ac4cc6df504ab0_l3.png "Rendered by QuickLaTeX.com")
![\[\left( 1,\text{ }6,\text{ }6 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ad7ebd045e8fb864994e31fedd35f402_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \text{ }4,\text{ }9,\text{ }6 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f084d6787d9e45924cee8137e66696c3_l3.png "Rendered by QuickLaTeX.com")
Solution: Let the points be x\[\left( 0,\text{ }7,\text{ }10 \right)\], y\[\left( 1,\text{ }6,\text{ }6 \right)\]& z\[\left( \text{ }4,\text{ }9,\text{ }6 \right)\] Firstly let us calculate the...
Solution: \[\left( \mathbf{0},\text{ }\mathbf{7},\text{- }\mathbf{10} \right)\] , \[\left( \mathbf{1},\text{ }\mathbf{6},\text{ }\text{ -}\mathbf{6} \right)\]and \[\left( \mathbf{4},\text{...
![\[\left( \mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{5} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-64cf62a3c8e5b50193da40694298234e_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{7},\text{ }\mathbf{0},\text{ }\mathbf{1} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-683a7c8881bc5cac555bfc80faf748c2_l3.png "Rendered by QuickLaTeX.com")
Solution: If three points are collinear, then they lie on a line. Firstly let us calculate distance between the 3 points i.e. AB, BC and AC Calculating AB A ≡ \[\left( \mathbf{2},\text{...
![\[\left( 2,\text{ }1,\text{ }3 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8dd835560c638fc2443260f7fa76b4f4_l3.png "Rendered by QuickLaTeX.com")
Let P be \[\left( 2,\text{ }1,\text{ }3 \right)\]and y be \[\left( 2,\text{ }1,\text{ }3 \right)\] By using the formula, Distance Py = \[\sqrt{[{{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{...
![\[\left(- 1,\text{ }3,\text{ }\text{- }4 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8088405a2a26025c802dc0444041003e_l3.png "Rendered by QuickLaTeX.com")
![\[\left( 1,\text{ -}3,\text{ }4 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3f12268e52b00109c360f70df2161e2d_l3.png "Rendered by QuickLaTeX.com")
Let x be \[\left( -1,\text{ }3,\text{ }\text{- }4 \right)\]and y be \[\left( 1,\text{ -}3,\text{ }4 \right)\] By using the formula, Distance xy = \[\sqrt{[{{({{x}_{2}}~\text{...
![\[(-3,7,2)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-77c9efe33cd43f75480912a71fae2220_l3.png "Rendered by QuickLaTeX.com")
![\[(2,4,-1)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f0aea65f5d7b3b7ba981925466ac17b3_l3.png "Rendered by QuickLaTeX.com")
Let x be \[(-3,7,2)\] and y be \[(2,4,-1)\] By using the formula, Distance xy = \[\sqrt{[{{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{ }{{({{y}_{2}}~\text{ }{{y}_{1}})}^{2}}~+\text{...
![\[\left( \mathbf{4},\text{ }\mathbf{3},\text{ }\mathbf{1} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bdfbf6e273e47a8700b272c5da2e9a13_l3.png "Rendered by QuickLaTeX.com")
Given Let x be \[\left( \mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{5} \right)\]and y be \[\left( \mathbf{4},\text{ }\mathbf{3},\text{ }\mathbf{1} \right)\] Use the formula, Distance xy...
Solution: (i) The x-axis and y-axis taken together determine a plane known as XY Plane. (ii) The coordinates of points in the XY-plane are of the form \[(x,y,0)\] (iii) Coordinate planes divide the...
![\[(1,2,3)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9fda469d657d838533f8a372c2974cc1_l3.png "Rendered by QuickLaTeX.com")
![\[(4,-2,3)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6dee6aade234dd58f0d2b6c746bd28b2_l3.png "Rendered by QuickLaTeX.com")
![\[(4,-2,-5)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d1e8948b57bb02bb0da940f1dc09d80f_l3.png "Rendered by QuickLaTeX.com")
![\[(4,2,-5)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6d95c44559deac1c81850895e7fa2b34_l3.png "Rendered by QuickLaTeX.com")
![\[(-4,2,-5)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9125371f3c5a63776080db78df15d242_l3.png "Rendered by QuickLaTeX.com")
![\[(-4,2,5)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3884787fb79274f08ea0db741570536b_l3.png "Rendered by QuickLaTeX.com")
![\[(-3,-1,6)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-efabe428979ff721d5d5c760de1f28b7_l3.png "Rendered by QuickLaTeX.com")
Solution: The below table which represents the octants: (i) \[(1,2,3)\] Here x, y and z are positive. So it lies in I octant. (ii) \[(4,-2,3)\] Here x and z are positive, y is negative So it lies...
Solution: given a point is in XZ plane if a point is in XZ-plane then its y-co-ordinate is \[0\].
Solution: Given a point is on the x-axis, if a point is on the x-axis, then the coordinates of y and z are \[0\]. Therefore, the point is \[(x,0,0)\]